The third order modular linear differential equations
Masanobu Kaneko1)1 Kiyokazu Nagatomo2)2 and Yuichi Sakai3)
1) Faculty of Mathematics, Kyushu University Motooka 744, Nishi-ku, Fukuoka 819-0395, JAPAN
2) Department of Pure and Applied Mathematics Graduate School of Information Science and Technology
Osaka University, Toyonaka, Osaka 560-0043, JAPAN
3) Yokomizo 3012-2, Oki-machi, Mizuma–gun, Fukuoka 830-0405, JAPAN
Abstract
We propose a third order generalization of the Kaneko-Zagier modular differential equa- tion, which has two parameters. We describe modular and quasimodular solutions of integral weight in the case where one of the exponents at infinity is a multiple root of the indicial equa- tion. We also classify solutions of “character type”, which are the ones that are expected to relate to characters of simple modules of vertex operator algebras and one-point functions of two-dimensional conformal field theories. Several connections to generalized hypergeometric series are also discussed.
AMS Subject Classification 2010: Primary 11F11, 81T40, Secondary 17B69
Key words: Vertex operator algebra, Modular invariance, Modular linear differential equa- tion,n-dimensional conformal field theory
1 Introduction
This paper studies a third order generalization of the Kaneko-Zagier equation (K-Z equation for short), which is called thethird order K-Z equation here. The K-Z equation first appeared in [11] in connection with supersingular j-invariants of elliptic curves, and subsequently, various modular and quasimodular solutions of the K-Z equation were found and studied in [5]–[8], etc. One of the characteristic properties of the K-Z equation is the invariance of the space of solutions under the standard slash action of the modular group Γ1 = SL2(Z), and indeed, our generalization has the same property.
1This work was supported by Japan Society for the Promotion of Science, Grant-in-Aid for Scientific Research (B) 23340010.
2This work was supported by Japan Society for the Promotion of Science, Grant-in-Aid for Challenging Ex- ploratory Research 26610007. The second author was also partially supported by Max-Planck Institute f¨ur Math- ematik.
Under mild conditions on the coefficient functions, we determine in §2 the form of what we call the third ordermodular linear differential equationas
f′′′−k+ 2
4 E2(τ)f′′+
{(k+ 1)(k+ 2)
4 E′2(τ) +αE4(τ) }
f′(τ)
−
{k(k+ 1)(k+ 2)
24 E2′′(τ) +kα
4 E4′(τ)−βE6(τ) }
f(τ) = 0, (1) where τ is a variable in the complex upper half-plane H, and ′ is the Euler operator of q(= e2π√−1τ) (see Theorem 1 in §2). The Ek(τ) is the normalized Eisenstein series of weight kgiven by
Ek(τ) = 1− 2k Bk
∑∞ n=1
σk−1(n)qn,
where Bk is the kth Bernoulli number and σm(n) is the sum of mth powers of positive divisors ofn. The parameterkis expected to stand for the weight off, andα, β are complex parameters.
As shown in [8], the K-Z equation is closely related to two-dimensional conformal field theory (2DCFT for short). The papers [1] and [3], which may be viewed as companion papers of the present one, study affine 2DCFT with at most 20 simple modules or 5 independent pseudo-characters and the minimal models with at most three simple modules, respectively.
The (formal) characters of such 2DCFT were expected to satisfy one of the third order K-Z equations.
One of the main results in this paper (given in §3) is an almost complete description of modular and quasimodular solutions in the case where the indicial equation of (1) withβ = 0 atq = 0 has a multiple root andkis an integer (Theorem 2 in§3.1, Theorem 4 in§3.2). The other is the determination of solutions ofcharacter type, which is characterized by integrality and positivity of Fourier coefficients of an associated weight 0 function (Proposition 6 and Theorem 7 in§4). As motivated by [10], we discuss in§5 a relation between the third order K-Z equation and hypergeometric series (Theorem 8 in§5).
2 Modular linear differential equations of third or- der
LetHbe the complex upper half-plane andF the space of holomorphic functions onH. The slash operator of weightk on F is defined as usual by
(f|kγ)(τ) = (cτ +d)−kf(γ(τ)) for eachγ =
(a b c d )
∈SL2(R)
for each real number k. In this section, we determine the general form of third order linear differential equations
f′′′(τ) +A(τ)f′′(τ) +B(τ)f′(τ) +C(τ)f(τ) = 0, ′ = q d
dq = 1 2π√
−1 d
dτ (2)
under the conditions that each coefficient A(τ), B(τ) and C(τ) are holomorphic on H, bounded as Im(τ) → ∞, and the space of solutions is invariant under the slash action |kγ for every γ ∈ Γ1 = SL2(Z). In general a linear ordinary differential equation on H with meromorphic coefficients is called amodular linear differential equation(MLDE) of weightk on a discrete subgroup Γ of SL2(R) if the space of solutions is invariant under the slash action|kγ of a fixed weight kfor each γ ∈Γ.
Theorem 1. Let f′′′(τ) +A(τ)f′′(τ) +B(τ)f′(τ) +C(τ)f(τ) = 0 be a third order linear differential equation for f such that the coefficient functions A(τ), B(τ) and C(τ) are holo- morphic on H and are bounded as Im(τ) → ∞. Then this is a modular linear differential equation of weight k onSL2(Z) if and only if the equation is given in the form
f′′′−k+ 2
4 E2(τ)f′′+
{(k+ 1)(k+ 2)
4 E′2(τ) +αE4(τ) }
f′(τ)
−
{k(k+ 1)(k+ 2)
24 E2′′(τ) +kα
4 E4′(τ)−βE6(τ) }
f(τ) = 0, (3) where α and β are complex numbers and ′=q d
dq.
Proof. The requirement that the equation is invariant under|kγ (γ ∈Γ1) implies (after a little complicated but similar calculations which were given in [6, Section 5] )
A
(aτ +b cτ +d
)
= (cτ +d)2A(τ)−3µ(k+ 2)(cτ +d), (4)
B
(aτ +b cτ +d
)
= µ(cτ +d)4B(τ)−2(k+ 1)(cτ +d)3A(τ) + 3µ2(k+ 1)(k+ 2)(cτ +d)2, (5) C
(aτ +b cτ +d
)
= (cτ +d)6C(τ)−kµ(cτ +d)5B(τ) +µ2k(k+ 1)(cτ+d)4A(τ)
−µ3k(k+ 1)(k+ 2)(cτ+d)3,
(6)
where
(a b c d )
∈ Γ1 and µ = c/2πi. By (4)–(6) and the transformation formula of the weight two (quasimodular) Eisenstein series
E2
(aτ+b cτ +d
)
= (cτ +d)2E2(τ) + 12µ(cτ+d) for any
(a b c d
)
∈Γ1, it follows that the three functions
A(τ) +(k+ 2)
4 E2(τ), B(τ) + (k+ 1)A′(τ), C(τ) +k
4B′(τ) + k(k+ 1)
12 A′′(τ) (7) are invariant under slash operators|2, |4 and |6 onγ ∈Γ1, respectively. BecauseA(τ),B(τ) andC(τ) are holomorphic and are bounded as Im(τ)→ ∞, these functions are holomorphic modular forms of weights 2, 4 and 6, respectively. By the well-known fact that there does not exist a non-zero holomorphic modular form of weight 2 on Γ1 and that the space of
holomorphic modular forms of weights 4 and 6 are both one-dimensional and are spanned by E4(τ) and E6(τ), respectively, we conclude that
A(τ) = −(k+ 2)
4 E2(τ), (8)
B(τ) = (k+ 1)(k+ 2)
4 E′2(τ) +αE4(τ), (9)
C(τ) = −k(k+ 1)(k+ 2)
24 E2′′(τ)−kα
4 E4′(τ) +βE6(τ) (10) with complex numbersα and β.
Conversely, the discussions above show that for any complex numbers α andβ, eq. (3) is a modular linear differential equation of weightk on Γ1.
We call (3) thethird order Kaneko-Zagier equation(of weightk) or thethird order K-Z equation (of weight k) for short.
Remarks. (1) Eq. (3) can be rewritten with the help of the Ramanujan relations 12E2′ = E22−E4, 3E4′ =E2E4−E6 and 2E6′ =E2E6−E42 as
ϑ3k(f) + ˆαE4ϑk(f) + ˆβE6(τ)(f) = 0 (11) with the values of ˆα=α−(3k2+ 12k+ 8)/144 and ˆβ =β+kα/12−k2(k+ 3)/864. Hereϑk(f) is the Serre derivativeϑk(f) :=f′−12kE2f and the iterated Serre derivations are defined by ϑ2k =ϑk+2◦ϑk and ϑ3k =ϑk+4◦ϑk+2◦ϑk.
(2) The third order modular differential equation obtained by applying the Serre derivation to the second order K-Z equation ϑ2k(f)− k(k+2)144 E4(f) = 0 is a special case of (11) with the values α= (k+ 1)(k+ 4)/72 and β= 0.
In the rest of the paper, we consider the special caseβ = 0 of (3), that is, the differential equation
f′′′(τ)−k+ 2
4 E2(τ)f′′(τ) +
{(k+ 1)(k+ 2)
4 E2′(τ) +αE4(τ) }
f′(τ)
−
{k(k+ 1)(k+ 2)
24 E2′′(τ) +kα 4 E4′(τ)
}
f(τ) = 0. (12) The reason why we restrict ourselves to this case is first to reduce the number of parameters from 3 to more manageable 2, but why we chooseβ = 0 among other specializations comes from our interest in 2DCFT or VOA. There, characters often take the formf /η2k wheref is a modular form of weightk= half of the central charge and is 1 +O(q) satisfying a modular differential equation, and η = q1/24∏∞
n=1(1−qn) is the Dedekind eta function. A typical example is the character of a lattice vertex operator algebra, which is written as ΘL/η2d, where L is a d-dimensional lattice and ΘL is the theta series associated to L. If f is a solution of (3) and f = 1 +O(q), then β should necessarily equal 0.
Our aim in the following two sections is to give a fairly complete description of both general and “character type” solutions of (3) under the assumption that the indicial equation
λ3−k+ 2
4 λ2+αλ = 0 (13)
has a multiple root and the weight kis an integer.
3 Third order Kaneko-Zagier equations with mul- tiple exponents
In this section we sutudy modular or quasimodular solutions of (12) when equation (13) has a multiple root and the weight parameter k is an integer. We are not able to establish any general statement when k is not an integer. However, computer experiments suggest that there are no (quasi)modular solutions (whose Fourier coefficients have bounded denominators) to (12) when equation (13) has a multiple root andk is not an integer.
Equation (13) has a triple root if and only if k = −2 and α = 0 (the root is 0). This case is trivial because (12) becomes f′′′ = 0 and a fundamental system of solutions of this equation is{1, τ, τ2}. We therefore assume in the remaining sections that (k, α)̸= (−2,0).
Clearly, a possible double root of (13) is eitherλ= 0 orλ= (k+ 2)/8. Ifλ= 0 is a double root, then α= 0 (andk̸=−2) and the corresponding MLDE is
f′′′−k+ 2
4 E2f′′+ (k+ 1)(k+ 2)
4 E2′f′−k(k+ 1)(k+ 2)
24 E2′′f = 0. (14) If λ = (k+ 2)/8 is a double root (k ̸= −2), then α = (k+ 2)2/64 and the corresponding MLDE is
f′′′−k+ 2
4 E2f′′+
{(k+ 1)(k+ 2)
4 E2′ +(k+ 2)2 64 E4
} f′
−
{k(k+ 1)(k+ 2)
24 E2′′+k(k+ 2)2 256 E4′
}
f = 0. (15) Convention. We sometimes use equation numbers in the text such as (14)k to make the dependence on the parameter kexplicit.
In fact, solutions of one of (14) and (15) are obtained from those of the other. More precisely, as is checked easily by direct computations, if f is a solution of (15)−2k−6 (resp.
of (14)−(k+6)/2), thenf∆(k+2)/4 (resp. f∆(k+2)/8) is a solution of (14)k (resp. of (15)k), and this correspondences is a bijection of the sets of solutions of (14) and (15). Therefore, we only need to consider either of equations (14) and (15). Or alternatively, by the equivalence
k >−2⇔ −2k−6<−2⇔ −(k+ 6)/2<−2,
it suffices to consider both (14) and (15) under the assumptionk >−2. Since we are mainly interested in (quasi)modular forms of positive weights, we hereafter assume k > −2 and study (14) and (15) separately in the following subsections.
3.1 Case λ = 0 is a double root
We describe solutions of (14)k when k is an integer > −2. If k = −1, (14)k becomes f′′′−14E2f′′= 0, and a set of fundamental solutions is given by 1, logq= 2π√
−1τ, and the double Eichler integral ofη6 (a solution of f′′=η6).
We define a sequence of functionsfk by the four-term recursion formula fk+4=akE4fk−bkE42fk−4+ck∆fk−8 (k≥4), where coefficients and initial values are given by
ak =
2(k3+k2−6k−12)
(k−3)(k+ 2)(k+ 4) ifk̸≡2 (mod 4), (k+ 6)2(k3+k2−6k−12)
128(k−3)(k+ 3)(k+ 4)2(k+ 5) ifk≡2 (mod 4), bk =
(k−2)k(k+ 1)
(k−3)(k+ 2)(k+ 4) ifk̸≡2 (mod 4), (k−2)2(k+ 2)2(k+ 6)2
65536(k−3)(k−1)(k+ 3)(k+ 4)2(k+ 5) ifk≡2 (mod 4), ck =
256(k−5)(k−4)k(k+ 1)
(k−6)(k−2)(k+ 2)(k+ 4) ifk̸≡2 (mod 4), bk ifk≡2 (mod 4),
and f−4 = 0, f−3 = η−6, f−2 = f−1 = f0 = 1, f1 = E2, f2 = −E2′/24, f3 = f4 = E4, f5 = E6+ 49E4′/15, f6 =−(21E4′′+ 10E6′)/151200, f7=E42+ 16E6′/105.
When k≡0,2 (mod 4), thesefks are same (up to normalization constants) as functions already given in [10, Theorem 3.1] and [7, Theorem 3.1]. We now have:
Theorem 2. Let k be a non-negative integer. The function fk is a solution of (14)k. Its q-expansion is fk = 1 +O(q) if k ̸≡ 2 (mod 4) and fk = q(k+2)/4 +O(q(k+6)/4) if k ≡ 2 (mod 4). Moreover,
(1) ifk≡0 (mod 4), fk is a modular form of weightk on SL2(Z),
(2) ifk≡2 (mod 4), fk is a quasimodular form of weight k+ 2 and depth 2 onSL2(Z), (3) ifk is odd, fk is a quasimodular form of weightk+ 1 and depth at most 1 onSL2(Z).
We can prove the theorem in a similar manner as in the proof of [7, Theorem 3.1] by using Lemma 3 below.
Let [f, g](k,ℓ)n (n≥0) be theRankin-Cohen bracket which is defined by [f, g](k,ℓ)n = ∑
r, s≥0 r+s=n
(−1)r
(n+k−1 s
)(n+ℓ−1 r
)
f(r)g(s), f(r) = (
q d dq
)r
f
for modular formsf andg of weightkandℓon Γ1 (indeed, the Rankin-Cohen bracket is well defined for any holomorphic functionsf and g), respectively. Then it is known that [f, g](k,ℓ)n is a modular form of weightk+ℓ+ 2non Γ1. An easy calculation shows that
[f, g](k,ℓ)1 = kf ϑℓ(g)−ℓϑk(f)g , [f, E4](k,4)2 = 10E4ϑ2k(f) +5
3(k+ 1)E6ϑk(f) + 5
72k2E42f , (16) [f, E6](k,6)2 = 21E6ϑ2k(f) +7
2(k+ 1)E42ϑk(f) + 7
48k2E4E6f . (17)
Lemma 3. Let k be a rational number.
(1) Suppose thatf is is a solution of(14)k. Then we have ϑk+8
(
[f, E4](k,4)2 )
= 5
63(k−1)[f, E6](k,6)2 , ϑk+10
(
[f, E6](k,6)2 )
= 7
20(k−2)E4[f, E4](k,4)2 −42k2(k+ 1)∆f , and [f, E4](k,4)2 /∆is a solution of (14)k−4.
(2)Suppose that Fk,Fk−4 andFk−8 are solutions of(14)k, (14)k−4 and(14)k−8, respectively.
ThenFk+4=E4Fk+E42Fk−4+ ∆Fk−8 is a solution of(14)k+4 if and only if [Fk, E6](k,6)2 + 2E4[Fk−4, E6](k2 −4,6)
=−7 8∆(
27648ϑk−4(Fk−4) + 12kE4ϑk−8(Fk−8) + (k2−2k+ 12)E6Fk−8)
. (18) Proof. We give a sketch of a proof. A fairly direct calculation of both sides by using equa- tion (11)k (withα=β= 0)
ϑ3k(f)(τ)−3k2+ 12k+ 8
144 E4(τ)ϑk(f)(τ)−k2(k+ 3)
864 E6(τ)f(τ) = 0 (19) together with (16) and (17) provides the equalities in (1).
That the [f, E4](k,4)2 /∆ is a solution of (14)k−4 can be seen by substituting [f, E4](k,4)2 /∆
into (19)k−4 and using the relation ϑk−4(
[f, E4](k,4)2 /∆)
=ϑk+8(
[f, E4](k,4)2 )
/∆ with (16).
Finally, substitutingFk+4 into (19)k+4, we see that the left-hand side of (19)k+4 coincides with
− 1 21
(
[Fk, E6](k,6)2 + 2E4[Fk−4, E6](k2 −4,6) +7
8∆(
27648ϑk−4(Fk−4) + 12kE4ϑk−8(Fk−8) + (k2−2k+ 12)E6Fk−8
)),
which proves (3).
The condition (18) in (2) of the lemma is effectively used in proving that the recursively defined functionfk satisfies the differential equation (14)k.
3.2 Case λ = (k + 2)/8 is a double root
Solutions of (15)k are described similarly as in the previous subsection. However, we need modular forms of level 2 here, and computer experiments suggest that there exist modular or quasimodular solutions only whenkis even.
Let H2(τ) = 2E2(2τ)−E2(τ) and ∆2(τ) = η(2τ)8/η(τ)4, which are modular forms of weight 2 on Γ0(2) and Γ(2), respectively. The groups Γ0(2) and Γ(2) are the standard
congruence subgroups of SL2(Z) of level 2;
Γ0(2) =
{(a b c d
)
∈SL2(Z)
c≡0 mod 2 }
, Γ(2) =
{(a b c d
)
∈Γ0(2) (
a b c d
)
≡ (1 0
0 1 )
mod 2 }
.
For even integers k̸≡6 (mod 8), we define gk by the four-term recursion formula gk+8=E42gk−bkE4∆gk−8+ck∆2gk−16 (k≥8)
with coefficients
bk= 512k(k3−8k2−4k+ 128)
(k−6)2(k+ 2)2 , ck= 65536(k−12)(k−10)2(k−8)k(k+ 4) (k−14)2(k−6)2(k+ 2)2
and with initial values g−8 = g−6 = g−4 = 0, g0 = 1, g2 = H2, g4 = E4, g8 = E42, g10 = H2(H24+ 61440∆42), g12= (3E43−2048∆)/3.
We define another series of functions hk for even k≡2 (mod 4) by the similar recursion hk+8=akE42hk−bkE4∆hk−8+ck∆2hk−16 (k≥10)
with coefficients
ak = (k+ 2)2(k+ 10)
256(k+ 4)(k+ 6)(k+ 8), bk = (k+ 2)(k+ 10)(k3−8k2−4k+ 128) 128(k−4)(k−2)(k+ 4)(k+ 6)(k+ 8), ck = (k−10)(k−6)(k+ 2)(k+ 10)
256(k−4)(k−2)(k+ 6)(k+ 8)
and initial values h−6 = η−12, h−2 = 1, h2 = ∆2, h6 = E4′′/240, h10 = ∆32(H22+ 192∆22/5), h14= (E43−720∆)′′/786240. Then we have:
Theorem 4. Let k be a non-negative even integer. The functions gk for k̸≡6 (mod 8)and hk fork≡2 (mod 4)are solutions of(15)kwith Fourier expansions of the formgk= 1+O(q) and hk=q(k+2)/8+O(q(k+10)/8). Moreover,
(1) ifk≡0 (mod 4), the function gk is a modular form of weight k onSL2(Z),
(2) if k ≡ 2 (mod 8), the function gk and hk are modular forms of weight k on Γ0(2) and Γ(2), respectively,
(3) if k≡ 6 (mod 8), the function hk is a quasimodular form of weight k+ 2 and depth at most 2 onSL2(Z) .
Remarks. (1) Since c8 =c10 =c12 = 0, we may choose any functions as the initial values g−8, g−6, and g−4. The following (non-modular) functions are solutions of (15)−8, (15)−6, and (15)−4 with 1 +O(q) respectively.
g−8 = 9 16η18
((
log(q)E2+ 12) ∫ q
0
E2∆3/4 E42
dq q −E2
∫ q
0
(log(q)E2+ 12)
∆3/4 E42
dq q
) , g−6 = 1
2η12
∫ q
0
∆2dq
q , g−4 = 1 16η6
∫ q
0
∫ q
0
η6 (dq
q )2
.
(2) From the general theory of ordinary differential equations, we know that if f and g are two independent solutions of (3)k, then the other (meromorphic) solution is given by
g
∫ q
0
∆k+24 f3{
(g/f)′}2
dq q −f
∫ q
0
g∆k+24 f4{
(g/f)′}2
dq q .
The proof of Theorem 4 goes similarly to that of Theorem 2 if we replace Lemma 3 by the lemma below.
Lemma 5. Let k be a rational number.
(1) Suppose thatf is a solution of (15)k. Then we have ϑk+8(
[f, E4](k,4)2 )
= 5
63(k−1)[f, E6](k,6)2 + 5
128(k+ 2)2E4[f, E4](k,4)1 , ϑk+10(
[f, E6](k,6)2 )
= 7
20(k−2)E4[f, E4](k,4)2 +21
64(k+ 2)2E6[f, E4](k,4)1 −42k2(1 +k)∆f , and (
[f, E4](k,4)2 +185(k+ 1)[f, E6](k,6)1 )
/E4∆is a solution of (15)k−8.
(2)Suppose thatFk,Fk−8andFk−16are solutions of(15)k,(15)k−8and(15)k−16, respectively.
ThenFk+8=E42Fk+E4∆Fk−8+ ∆2Fk−16 is a solution of (15)k+8 if and only if 1
5E6[Fk, E4](k,4)2 +k+ 6
48 [Fk, E43](k,12)1 + 576(k+ 3)∆ϑk(Fk) + ∆
( 1
21[Fk−8, E6](k2 −8,6)+ 1
24[Fk−8, E42](k1 −8,8)−9k2+ 36k+ 292
576 E4E6Fk−8 )
= ∆2 (
−k−2
4 E4ϑk−16(Fk−16) +k2+ 44
96 E6Fk−16 )
.
3.3 Quasimodular forms and solutions with logarithmic terms
We found solutions of quasimodular forms of the third order K-Z equations in Theorems 2 and 4. A simple observation shows that, because of the modular invariance of the space of solutions, if MLDEs of weight k have solutions of quasimodular forms of weight k+r and depth r >0, then there exist solutions with logarithmic terms. We briefly illustrate this in the cases of depth 1 and 2.
Suppose that a quasimodular form Jk := AkE2 +Bk of weight k+ 1 and depth 1 is a solution of a K-Z equation of order 3 and weight k, where Ak and Bk are modular forms on Γ1 of weight k−1 and k+ 1, respectively. Then the function Gk := τ−kJk(−1/τ) = (2π√
−1)−1(Jklogq+ 12Ak) is a solution because of the modular invariance. Moreover, it follows from the transformation formula of E2 that
(Jk Gk
)
k
(a b c d
)
=
(d c b a
) (Jk Gk
)
for each
(a b c d
)
∈Γ1.
For example, eqs. (14)1 and (14)5 have solutions J1 =E2 and J5 = (49E2E4−4E6)/45, respectively. ThenG1= (2π√
−1)−1(J1logq+12) andG5= (2π√
−1)−1(F5logq+196E4/15) are solutions of (14)1 and (14)5 with logarithmic terms, respectively.
Next suppose that a quasimodular form Kk :=AkE22 +BkE2+Ck of weight k+ 2 and depth 2 is a solution of a third order K-Z equation of weight k, where Ak, Bk and Ck are modular forms on Γ1 of weightk−2,kandk+ 2, respectively. Then we have solutions with logarithmic terms
Ik = (2π√
−1)−2(
Kk(logq)2+ 12(2AkE2+Bk) logq+ 144Ak
) , Gk = (2π√
−1)−1(Kklogq+ 6(2AkE2+Bk)),
which are obtained by Ik = Kk|kγ1 and Gk = (Kk|kγ2 −Kk−Ik)/2 (γ1 :τ 7→ −1/τ, γ2 : τ 7→ −1/(τ+ 1)), respectively. It then follows that
Kk Gk Ik
k
(a b c d )
=
d2 2cd c2 bd ad+bc ac b2 2ab a2
Kk Gk Ik
for each
(a b c d
)
∈Γ1.
For instance, eq. (15)6 has a solution K6 = (E22E4 −2E2E6 +E42)/1728. Consequently, there exist solutionsI6 = (2π√
−1)−2(
K6(logq)2+ (E2E4−E6) logq/72 +E4/12)
andG6 = (2π√
−1)−1(K6logq+ (E2E4−E6)/144).
This kind of phenomena that if a MLDE has a quasimodular solution of depth r then associated vector-valued modular form corresponds to the symmetric tensor representation with degreer+ 1 is studied for instance in [4] and [9].
4 Solutions of character type
As mentioned in the introduction, we are interested in special type of solutions of MLDEs in connection with 2DCFT and VOAs, namely, solutions of “character type” which is defined as follows.
Definition. A solution f of the third order K-Z equation (of weight k) is said to be of character typeif it is a (quasi)modular form and all Fourier coefficients of f /η2k are non- negative integers. Furthermore, if its leading Fourier coefficient is 1, we call it of vacuum character type.
In this section we shall give a fairly complete description of solutions of character type of equation (14). As explained in the paragraph before §3.1, equations (14) and (15) are equivalent in the sense that solutions of either of these equations are obtained from the other by multiplying a suitable power ofη. And under this equivalence, the property of a solution being of character type is clearly unchanged because we look at the Fourier coefficients of the associated weight 0 function obtained by multiplying a power of η. Hence, we shall exclusively look at equation (14), but for any (positive and non-positive) integer k.
Let f be a solution of (14)k and setg=f /η2k. Then (14)k is rewritten in terms ofg as g′′′−1
2E2g′′+ {1
2E2′ −k(k+ 4) 48 E4
}
g′−k2(k+ 3)
864 E6g = 0. (20)
Our aim in this section is to determine all solutionsg of (20)k which have the form g = qν
( 1 +
∑∞ n=1
anqn )
, (21)
where eachanis anon-negative integer andν ∈R. Since the characteristic equation of (20) is
ν3−1
2ν2−k(k+ 4)
48 ν−k2(k+ 3)
864 =
( ν+ k
12 )2(
ν− k+ 3 6
)
= 0,
the indicial roots of (20) are {−k/12,−k/12,(k+ 3)/6}. The indicial roots ν =−k/12 and (k+3)/6 ofgcorrespond to the exponentsµ= 0 and (k+2)/4 off, respectively. We compute coefficients an of a solution (21) of (20) by the Frobenius method, and seek for conditions thatanare non-negative integers. (Even if an index is a double root, we can obtain a solution when a pair of indices does not have a integral difference.)
Substituting (21) into (20), we have 2
(
n+ν+ k 12
)2(
n+ν−k+ 3 6
) an =
∑n i=1
(
n+ν−i)2e2,i−(
i·e2,i−k(k+ 4) 24 e4,i
)
(n+ν−i) +k2(k+ 3) 432 e6,i
)
an−i (22) with a0 = 1, where E2(τ) = ∑∞
i=0e2,iqi, E4(τ) = ∑∞
i=0e4,iqi and E6(τ) = ∑∞
i=0e6,iqi, respectively.
We consider two cases, that is, ν=−k/12 andν = (k+ 3)/6, separately in the following subsections.
4.1 Case ν = − k/12
We now study the solutions of (20)k with an indicial root ν =−k/12, which corresponds to the solutions of (14)kwith an exponentµ= 0. Though we are not able to prove the positivity of Fourier coefficients for all weights, we can list all possible solutions of vacuum character type when the weights are integers.
Theorem 6. Let k be an integer.
(1) Suppose that the equation (14)k has a solution of vacuum character type with the expo- nent 0. Then k is one of the values in the set
{ −30,−22,−14,−10,−6,−4,0,3,4,8}. (23)
(2)For each k∈ {−10,−6,−4,0,3,4,8}, the functionvk given below is a solution of (14)k
of vacuum character type with the exponent0 ; v0 = 1,
v3 = v4 = E4 = 1 + 240q+ 2160q2+ 6720q3+ 17520q4+ 30240q5+· · · , v8 = E42 = 1 + 480q+ 61920q2+ 1050240q3+ 7926240q4+ 37500480q5+· · ·, v−4 = ∆2
√∆ = 1 + 16q+ 144q2+ 960q3+ 5264q4+ 25056q5+· · ·, v−6 = h6
∆ = 1 + 60q+ 1440q2+ 22080q3+ 253680q4+ 2369160q5+· · · , v−10 = h14
∆2 = 1 + 240q+ 18540q2+ 792960q3+ 23080560q4+ 508465440q5+· · · , where the subscripts indicate the weights of the solutions.
(3)For eachk=−14,−22,−30, we have the following solutionvk of(14)k, which is possibly of vacuum character type;
v−14 = h22
∆3 = 1 + 546q+ 88452q2+ 7440888q3+ 405394080q4+ 16071109236q5+· · ·, v−22 = h38
∆5 = 1 + 1540q+ 657360q2+ 137466120q3+ 17723389420q4+· · ·, v−30 = h54
∆7 = 1 + 3045q+ 2494870q2+ 974923740q3+ 229294066260q4+· · · , where ∆2(τ) =η(2τ)8/η(τ)4 andhk is a quasimodular form defined in §3.2.
(4) The functions vk (k = 0,3,4,8) are modular forms of weight k (except that v3 has weight4)and vk (k=−6,−10,−14,−22,−30)are quasimodular forms of weight k+ 2and depth2 onSL2(Z). The functionv−4 is a meromorphic modular form on Γ0(2) of weight−4 with the pole at the cusp 0.
Proof. Using the recursive formula (22) withν =−k/12 2n2
(
n−k+ 2 4
) an
=
∑n i=1
(
(n− k
12 −i)2e2,i−(
i e2,i−k(k+ 4) 24 e4,i
)(
n− k 12 −i
)
+ k2(k+ 3) 432 e6,i
) an−i,
(24) we can determine the Fourier coefficientsan of a solutiong of (20) of the form (21).
The case n = 1 of (24) gives (1−k/2)a1 = −2k3 −7k2 −2k. Then it follows that a1 = 4k2+ 22k+ 48 + 96/(k−2). Suppose thata1is an integer. Then the condition (k−2)|96 givesk−2 =±1,±2,±3,±4,±6,±8,±12,±16,±24,±32,±48,±96. The another condition a1 ≥0 reduces the values of kto
{−94,−46,−30,−22,−14,−10,−6,−4,0,3,4,5,6,8,10,14,18,26,34,50,98}. (25) We pick up each value ofkfrom the list (25) and determine everyanby (24) recursively (up to n= 30). Then the list ofksuch that all anare non-negative integers for 0< n≤30 reduces
the values ofkto{ −30,−22,−14,−10,−6,−4,0,3,4,8}, which coincides with (23). For eachk in (23), it follows from Theorem 2 and Theorem 4 that the functionsvk are solutions of (14)k.
For k = 0,3,4 and 8, the function vk is clearly of vacuum character type because E4
and 1/η2k have positive integral Fourier coefficients. Since v−4η8 =q1/3∏∞
n=1(1 +qn)8 and v−6η12 = E4′′/(240η12), the functionsv−4 and v−6 are also of vacuum character type.
For v−10, we use the identity v−10η20 = 1
12η28 (
5E4 (E4′
240 )2
+ 7 (E6′
504 )2)
= 1
12η28 (
5 (
1 + 240
∑∞ n=1
σ3(n)qn )(∑∞
n=1
σ3(n)qn )2
+ 7 (∑∞
n=1
σ5(n)qn )2)
which shows the positivity of Fourier coefficients. Moreover, using the congruence σ3(n) ≡ σ5(n) (mod 12), we have
5 (
1 + 240
∑∞ n=1
σ3(n)qn )(∑∞
n=1
σ3(n)qn )2
+ 7(∑∞
n=1
σ5(n)qn)2
≡5 (∑∞
n=1
σ3(n)qn )2
+ 7 (∑∞
n=1
σ5(n)qn )2
≡12 (∑∞
n=1
σ3(n)qn )2
≡ 0 (mod 12) and thus conclude that the Fourier coefficients of v−10η20 are integers.
Because ∆2/√
∆ = (H22−64∆22)−1 and ∆22 is a modular form on Γ0(2), the functionv−4
is a modular form on Γ0(2) of weight−4. Since
∆2(τ)
√∆(τ) −4
(0 −1
1 0
)
= η(τ /2)8 η(τ)16 = 1
q1/2 −8 + 36q1/2+O(q), (26) v−4 has a pole at 0. The rest of assertions are clear from the definition of the functions vk.
Remark. It is worth noting that the solutionv−10 has an another expressionv−10= (E43− 720∆)′′/(786240∆2), and the modular form E43 −720∆ of weight 12 is the theta series of the 24-dimensional Leech lattice. Let Nm be the number of vectors of the Leech lattice whose norms are m. The identity
E43−720∆ = E12− 65520
691 ∆ = 1 + 65520 691
∑∞ n=1
(σ11(n)−τ(n)) qn, whereτ(n) is the nth Fourier coefficient of ∆(τ), shows the well-known formula
N2n = 65520
691 (σ11(n)−τ(n)). (27)
By the integrality of coefficients of (E34−720∆)′′/786240 proved in the previous paragraph, we derive the fact that (i)N2n is divisible by N4 = 196560 ifn is even, (ii) N2n is divisible
by 4N4 = 786240 if n is odd and not divisible by 3 and (iii) 3N2n is divisible by 4N4 if nis odd and divisible by 3 since 786240 = 65520×12 = 196560×3. (The facts (i)–(iii) can also be proved by using (27) and classical congruences ofτ(n) given in [13].)
Example. The function v8/η16 = j2/3 is the character of the lattice VOA associated with the unimodular lattice of rank 8. Herej(τ) =E4(τ)3/∆(τ) is the elliptic modular function.
At the present time, we do not know if vk is of character type except a few k. However, computer experiments suggest that any solution of (14)k with the exponent 0 is of character type after multiplied by a suitable integer. More specifically, we give the following conjecture.
Conjecture. Letk be either a positive integer or a negative even integer. Define a natural numberp1(k) by
p1(k) =
⌊(k+1)/8∏ ⌋ i=1
(k−4i+ 2) ifk is positive,
⌊(|k|−∏4)/4⌋ i=1
(|k|/2 + 2i−1) ifk is negative,
where an empty product is regarded as 1. Ifvk= 1 +O(q) is a solution of (14)k, thenp1(k)vk is of character type.
For example, we have p1(7) = 5 and
5v7= 5E42+ 16E6′/21 = 5 + 96
∑∞ n=1
(25σ7(n)−nσ5(n))qn.
Since 25σ7(n)−4nσ5(n) >0 (n≥1) (this is because σ7(n)≥n7 ≥n(15+ 25+· · ·+n5) ≥ nσ5(n)), we see that p1(7)·v7 is a solution of (15)7 of character type.
4.2 Case ν = (k + 3)/6
We next study the solutions of (20) with index ν = (k+ 3)/6, which corresponds to those of (14) with exponent µ= (k+ 2)/4.
Theorem 7. Let k be an integer.
(a) Suppose that(14)k has a solution of vacuum character type with the exponent (k+ 2)/4.
Thenk is one of the values in the list
{ −8,−7,−5,−4,2,6}. (28)
(b) For each k in (28), the following functions wk are of vacuum character type with the
